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Integral of csc2 is -ctn

by himanshu121
Tags: csc2, integral, integration
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himanshu121
#1
Dec10-03, 09:34 AM
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Here is the Problem:

[tex] \int_0^\pi\theta^2cosec^2\theta d\theta[/tex]


I want to do this problem by parts and finding hard to find shortest way to do the problem any shortest way pls
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mathman
#2
Dec10-03, 04:45 PM
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The straightforward way is probably the easiest. integral of csc2 is -ctn and integral of ctn is log(sin). Integrating log(sin) may take a little work.
PrudensOptimus
#3
Dec10-03, 05:05 PM
P: 640
Originally posted by himanshu121
Here is the Problem:

[tex] \int_0^\pi\theta^2cosec^2\theta d\theta[/tex]


I want to do this problem by parts and finding hard to find shortest way to do the problem any shortest way pls
cos*sec^2(x) or cosx*sec^2(x)?

master_coda
#4
Dec10-03, 05:24 PM
P: 678
Integral of csc2 is -ctn

Originally posted by PrudensOptimus
cos*sec^2(x) or cosx*sec^2(x)?
Isn't it csc(x)?
himanshu121
#5
Dec10-03, 09:49 PM
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P: 658
It is cosecant(x) i.e csc(x)
Integrating log(sin) may take a little work
This is probably the easiest way of doing which i too have tried but it is not the shortest way

There are many ways of doing a problem i am looking for shortest way
Thnxs
master_coda
#6
Dec10-03, 10:39 PM
P: 678
How about:

[tex]
\begin{align*}
\int_0^\pi\theta^2\csc^2\theta\;d\theta
&=\int_0^1\theta^2\csc^2\theta\;d\theta+\int_1^\pi\theta^2\csc^2\theta\ ;d\theta \\
&>\int_0^1\theta^2\csc^2\theta\;d\theta+\int_1^\pi\csc^2\theta\;d\th eta \\
&=\int_0^1\theta^2\csc^2\theta\;d\theta+\left[-\cot\theta\right]_1^\pi
\end{align*}
[/tex]

Now since [itex]\lim\limits_{\theta\rightarrow\pi}(-\cot\theta)=+\infty[/itex] we know that the rightmost term diverges. Moreover, the left integral (the one from 0 to 1) is clearly positive. Thus the original integral clearly diverges.
jk7711
#7
Dec10-03, 11:31 PM
P: 3
master_coda:
the integral of x^2*(cscx)^2 from 1 to pi comes out to be about 4.2
the integral of (cscx)^2 from 1 to pi goes to inifinite. I believe to say the integral of x^2*(cscx)^2 from 1 to pi is greater than the integral of (cscx)^2 from 1 to pi since according to the values it came out as then the formula should be switched around and divergence would not be proved.

himanshu121:
Best way to do it would be to plug it into a calculator or look it up in a table. Otherwise you're probably stuck with integrating log(sin(x)).

jk
himanshu121
#8
Dec11-03, 12:35 AM
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P: 658
Best way to do it would be to plug it into a calculator or look it up in a table
We are not allowed to use calculator in India till we are undergraduate
I found the way but dont know whether it is shortest one or not but definetly i wont stuck at log(sinx)



[tex]

I = \int_0^\pi\theta^2\csc^2\theta d\theta [/tex]

[tex]I= \int_0^\pi\ (\pi-\theta)^2\csc^2\theta d\theta [/tex]

this gives

[tex] \pi\int_0^\pi\csc^2 \theta d\theta = 2\int_0^\pi\theta\csc^2\theta d\theta [/tex]

integrating by parts with one part as [tex]\theta[/tex] and other as [tex]\theta\csc^2\theta d\theta [/tex]

i will get[tex] \int_0^\pi \cot\theta d\theta = \log(csc\theta-cot\theta)
[/tex]

much easier than integrating log(sinx)

But another problem is how i will put the limits in cotx from 0 to pi in both cases it is infinity and i know there is no break in the function cotx b/w these points
Hurkyl
#9
Dec11-03, 07:09 AM
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I agree with master_coda...

the integral of x^2*(cscx)^2 from 1 to pi comes out to be about 4.2
Where did you come up with this?! My calculator says 3*10^14 with the caveat of "questionable accuracy" (though, 3*10^14 is a good approximation of infinity. )
chroot
#10
Dec11-03, 09:44 AM
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Originally posted by Hurkyl
(though, 3*10^14 is a good approximation of infinity. )
Yes, for large values of 3*10^14.

- Warren
master_coda
#11
Dec11-03, 09:45 AM
P: 678
I'm pretty sure that the integral diverges. My algebra, my calculator and my computer all agree.

Also, if it's any help, [itex]\csc\theta-\cot\theta=\frac{1-\cos\theta}{\sin\theta}[/itex].
jk7711
#12
Dec13-03, 09:02 PM
P: 3
hurkyl. yeah that 4.2 was kinda off huh? just keeping you on your toes i guess:P i entered it in wrong but i redid it and came up with what you got.

i used u=(x^2)csc^2(x) and dv=dx and came up with the first term (x^3*csc^2(x)) going to infinite so i think you might be right coda.

jk
harsh
#13
Dec17-03, 03:54 PM
P: 76
Why dont you try using the tabular method to doing this problem.
Since theta^2 will eventually go to zero if you keep on taking the derivatives, you should do it by tabular method


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